Idempotent Ring is Commutative

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Theorem

Let $\struct {R, +, \circ}$ be an idempotent ring.

Denote with $0_R$ the zero of $R$.


Then $\struct {R, +, \circ}$ is a commutative ring.


Proof

Let $x, y \in R$.

Then:

\(\displaystyle x + y\) \(=\) \(\displaystyle \paren {x + y}^2\) Definition of Idempotent Ring
\(\displaystyle \) \(=\) \(\displaystyle x^2 + x \circ y + y \circ x + y^2\) Binomial Theorem: Ring Theory
\(\displaystyle \) \(=\) \(\displaystyle x + x \circ y + y \circ x + y\) Definition of Idempotent Ring


Subtracting $x + y$ from both sides yields:

\(\displaystyle x \circ y + y \circ x\) \(=\) \(\displaystyle 0_R\)
\(\displaystyle \) \(=\) \(\displaystyle y \circ x + y \circ x\) Idempotent Ring has Characteristic Two


Finally, subtracting $y \circ x$ from both sides, we obtain:

$x \circ y = y \circ x$

and conclude $R$ is a commutative ring.

$\blacksquare$


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